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Alabama Instruments Company has set up a production line to manufacture a new calculator. The rate of production of these calculators after $ t $ weeks is$$ \frac{dx}{dt} = 5000 \biggl( 1 - \frac{100}{(t + 10)^2} \biggr) \text{calculators/week} $$ (Notice that production approaches 5000 per week as time goes on, but the initial production is lower because of the workers' unfamiliarity with the new techniques.) Find the number of calculators produced from the beginning of the third week to the end of the fourth week.
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02:56
Frank Lin
01:41
Amrita Bhasin
Calculus 1 / AB
Chapter 5
Integrals
Section 5
The Substitution Rule
Integration
Campbell University
Oregon State University
University of Nottingham
Lectures
05:53
In mathematics, an indefinite integral is an integral whose integrand is not known in terms of elementary functions. An indefinite integral is usually encountered when integrating functions that are not elementary functions themselves.
40:35
In mathematics, integration is one of the two main operations of calculus, with its inverse operation, differentiation, being the other. Given a function of a real variable (often called "the integrand"), an antiderivative is a function whose derivative is the given function. The area under a real-valued function of a real variable is the integral of the function, provided it is defined on a closed interval around a given point. It is a basic result of calculus that an antiderivative always exists, and is equal to the original function evaluated at the upper limit of integration.
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okay. And this problem x is the number of calculators produced by a calculator company. And then I've given dx DT, which is the rate at which the calculus leaders are being produced in calculators per week. So what they want to know is the number of calculators produced from the beginning of the third week until the end of the fourth week. So this is in calculators per week. So if I integrate with respect to time, then I will find out how Maney calculators they make. All right, eps All right. So the integral from the beginning of the third week. So from the third week until the end of the fourth week, So I'm gonna put 3 to 5, because if you look at it 012345 Well, we wanted to start here, and we wanted to stop here at the end of the fourth week. Okay, 5000 times one minus 100 over T plus two. That looks t plus 10 squared DT. Okay, so I know that I could move this 5000 out because it's just a constant, and then I can integrate the wand. But then I have a problem here with a T plus 10. So what I'm gonna do is make a substitution there. So I have 5000 3 to 5 one minus 100 T plus 10 to the minus two DT. All right, I'm gonna let you equal to t plus 10. And then do you is DT. And if t equals three u equals three three plus 10 years. 13 if t equals five, you goes five plus 10, which is 15. So this integral turns into 5000 13 to 15, one minus 100. You to the minus two, do you? And now I can integrate with respect to you. So I get 5000, and the integral of one is you minus 100. And the integral of you to the negative two is remember, add one you to the negative one. Over negative one from 13 to 15. Since I switched the end points, I don't have to switch the you back, but I do. You wanna do another simplified before I start looking in? I get u minus minus so plus 100 over you from 13 to 15. Okay, so that's 5000 times 15 plus 100 over 15, minus 13. Plus 100 over 13. All right, see what we get. Yeah. 15 plus 100. Divide by 15. Oh, no, I don't think that's right. Let me try again. Ah. 100 divided by 15 plus 15. So we have 5000 times 21.666667 minus 100. Divided by 13 plus 13, 20.692 30 Directory 308 don't, dude. So 5000 times 0.974 Some other stuff equals 4871. So from the beginning of the third week to the end of the fourth week, I think that they can make 4871 calculators.
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